The number of values of k for which the system of linear equations $$(k+2)x + 10y = k$$ and $$kx + (k+3)y = k - 1$$ has no solution is:

We have two linear equations in the variables $$x$$ and $$y$$:

$$ (k+2)\,x + 10\,y = k \qquad\qquad\qquad (1) $$ $$ k\,x + (k+3)\,y = k-1 \qquad\qquad (2) $$

To study the nature of the solutions, we recall the standard condition for a pair of linear equations

$$ a_1x + b_1y = c_1, \qquad a_2x + b_2y = c_2 $$

Lines are

• coincident (infinitely many solutions) when $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.$$
• parallel and distinct (no solution) when $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}.$$
• intersecting (a unique solution) when $$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}.$$

For our system we identify

$$ a_1 = k+2,\; b_1 = 10,\; c_1 = k, $$ $$ a_2 = k,\; \;\;\;\; b_2 = k+3,\; c_2 = k-1. $$

First we impose the condition for the lines to be parallel:

$$ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}. $$

Substituting the coefficients, we get

$$ \dfrac{k+2}{k} = \dfrac{10}{k+3}. $$

Now we cross-multiply:

$$ (k+2)(k+3) = 10k. $$

Expanding the left side,

$$ k^2 + 5k + 6 = 10k. $$

Bringing all terms to the left,

$$ k^2 + 5k + 6 - 10k = 0, $$ $$ k^2 - 5k + 6 = 0. $$

This is a quadratic equation. Factoring,

$$ (k-2)(k-3) = 0. $$

So

$$ k = 2 \quad \text{or} \quad k = 3. $$

These two values make the ratios of the coefficients equal, hence the lines are either coincident or parallel. To know whether there is actually no solution, we also compare the ratio of the constants:

$$ \dfrac{c_1}{c_2} = \dfrac{k}{k-1}. $$

We examine both values separately.

For $$k = 2$$:

$$ \dfrac{a_1}{a_2} = \dfrac{4}{2} = 2, \qquad \dfrac{b_1}{b_2} = \dfrac{10}{5} = 2, $$ $$ \dfrac{c_1}{c_2} = \dfrac{2}{1} = 2. $$

All three ratios are equal, so the two equations represent the same line. There are infinitely many solutions, not zero solutions.

For $$k = 3$$:

$$ \dfrac{a_1}{a_2} = \dfrac{5}{3}, \qquad \dfrac{b_1}{b_2} = \dfrac{10}{6} = \dfrac{5}{3}, $$ $$ \dfrac{c_1}{c_2} = \dfrac{3}{2}. $$

The first two ratios are equal, but

$$ \dfrac{c_1}{c_2} = \dfrac{3}{2} \neq \dfrac{5}{3}. $$

Therefore the lines are parallel and distinct, so the system is inconsistent and has no solution.

Only one value, namely $$k = 3,$$ satisfies the required condition.

Hence, the correct answer is Option A.